Problem 53 - Physics for Scientists and Engineers with Modern Physics

Question

(1I) A projectile is fired with an initial speed of 46.6 \(\mathrm{m} / \mathrm{s}\) at an angle of \(42.2^{\circ}\) above the horizontal on a long flat firing range. Determine \((a)\) the maximum height reached by the projectile, (b) the total time in the air, \((c)\) the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after firing.

Step-by-Step Solution

Answer

Max height: 31.6 m, Time in air: 6.3 s, Range: 219 m, Velocity at 1.5 s: 36.8 m/s.

Step 1: Determine Initial Velocity Components

First, calculate the horizontal and vertical components of the initial velocity using trigonometric functions. Given the initial speed \(v_0 = 46.6\, \text{m/s}\) and the angle \(\theta = 42.2^{\circ}\):\[v_{0x} = v_0 \cos(\theta) = 46.6 \cos(42.2^{\circ}) \v_{0y} = v_0 \sin(\theta) = 46.6 \sin(42.2^{\circ})\]Calculate these values to proceed to the next step.

Step 2: Calculate Maximum Height

To find the maximum height, use the vertical component \(v_{0y}\) and the formula related to vertical motion:\[v_y^2 = v_{0y}^2 - 2g h \]At maximum height, \(v_y = 0\). Solving for \(h\) gives:\[0 = v_{0y}^2 - 2gh \h = \frac{v_{0y}^2}{2g}\]where \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity.

Step 3: Calculate Total Time in the Air

The total time of flight can be found by doubling the time it takes to reach the maximum height (the symmetry of projectile motion). First, calculate the time to reach the maximum height:\[v_y = v_{0y} - gt \0 = v_{0y} - gt \Rightarrow t = \frac{v_{0y}}{g}\]Total time is:\[t_\text{total} = 2 \times \frac{v_{0y}}{g}\]

Step 4: Calculate Total Horizontal Distance (Range)

The range of the projectile is given by:\[R = v_{0x} \times t_\text{total}\]Use the previously calculated horizontal velocity \(v_{0x}\) and the total time of flight to find the range.

Step 5: Determine Velocity 1.50 s After Firing

To find the velocity of the projectile after 1.50 seconds, calculate the horizontal and vertical velocity components at that time. The horizontal velocity component remains unchanged:\[v_x = v_{0x}\]The vertical velocity component changes with time:\[v_y = v_{0y} - gt\]Substitute \(t = 1.50\, \text{s}\) to find \(v_y\). The magnitude of the velocity vector is:\[v = \sqrt{v_x^2 + v_y^2}\]And the direction \(\phi\) is given by:\[\phi = \tan^{-1}\left(\frac{v_y}{v_x}\right)\]

Key Concepts

Maximum Height

When discussing the maximum height in projectile motion, we are looking to find the peak point the object reaches. The maximum height occurs when the vertical component of the velocity (\(v_y\)) becomes zero. At this point, the projectile stops moving upward and is about to start descending.
To calculate this, we rely on the initial vertical velocity (\(v_{0y}\)) and the acceleration due to gravity (\(g = 9.8 \text{ m/s}^2\)). The key formula used is:

• \(v_y^2 = v_{0y}^2 - 2g h\)
• At max height, \(v_y = 0\).

By setting \(v_y\) to zero, we solve for height (\(h\)):\[h = \frac{v_{0y}^2}{2g}\]This gives us the maximum height.
This concept is crucial in understanding how projectiles move under the influence of gravity.

Total Time of Flight

The total time of flight is the duration for which the projectile is in the air. It hinges on how long it takes to ascend and descend, thanks to gravity's pull. The projectile's trajectory is symmetric, meaning the time to reach the peak height is equal to the time to descend back to the ground.
You can calculate the time to reach the peak using the equation:\[t = \frac{v_{0y}}{g}\]Since the ascent and descent times are equal, the total time in the air is twice this time:\[t_\text{total} = 2 \times \frac{v_{0y}}{g}\]Understanding this symmetry and using these equations help in determining how long a projectile stays airborne, which is critical for applications like ballistics and sports.

Range of Projectile

The range of a projectile is how far it travels horizontally before hitting the ground. This is a key point of interest in projectile motion problems. The range is affected by the initial speed, the angle of launch, and gravity.
Using the horizontal component of initial velocity (\(v_{0x}\)) and the total time of flight (\(t_\text{total}\)), the range (\(R\)) can be found with the formula:\[R = v_{0x} \times t_\text{total}\]The simplicity of the formula belies the dependencies on both the time in the air and how fast the projectile moves horizontally. This concept finds its relevance in fields like engineering, physics and even gaming.

Velocity Components

The velocity of a projectile is divided into horizontal and vertical components, which are foundational to understanding projectile motion.

• **Horizontal velocity (\(v_{0x}\)):** This remains constant throughout the projectile's flight because no horizontal forces act on it (assuming air resistance is negligible).
• **Vertical velocity (\(v_{0y}\)):** This changes due to gravity, decreasing as the projectile rises and increasing as it descends.

To find the components at any time (\(t\)), we use:

• The horizontal component remains:\(v_x = v_{0x}\).
• The vertical component is updated with time:\[v_y = v_{0y} - gt\]

These components combine to give the total velocity of the projectile at any point in time. Calculating these helps understand the projectile's motion path and how its speed and direction evolve over time.